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Find the sum of the series \(\frac{x^{2}}{1.2}-\frac{x^{3}}{2.3}+\frac{x^{4}}{3.4}-\ldots+(-1)^{n+1} \frac{x^{n+1}}{n(n+1)}+\ldots,|x|<1\)

Short Answer

Expert verified
Answer: When |x| < 1, the sum of the convergent alternating series is given by \(\frac{x^2}{1}\).

Step by step solution

01

Identify the general term

The general term of the series is given by the formula: \((-1)^{n+1} \frac{x^{n+1}}{n(n+1)}\)
02

Recognize the pattern in the series

Let's look at the series' first few terms: \(\frac{x^2}{1.2} - \frac{x^3}{2.3} + \frac{x^4}{3.4} - \frac{x^5}{4.5} + \ldots\) Notice that we can write the terms of the series as differences of consecutive functions: \(\left(\frac{x^2}{1} - \frac{x^3}{2}\right) - \left(\frac{x^3}{1} - \frac{x^4}{2}\right) + \left(\frac{x^4}{1} - \frac{x^5}{2}\right) - \ldots\)
03

Observe the series telescoping

The series is telescoping, which means that, after summing up to a certain term, some of the terms will cancel out and leave just a few terms remaining. In this case, adding up the first n terms will result in: \(\frac{x^2}{1} - \frac{x^{n+2}}{2n}\)
04

Find the limit of the sum

As the series converges when |x| < 1, we can now find the limit of the sum as n goes to infinity: \(\lim_{n \to \infty} (\frac{x^2}{1} - \frac{x^{n+2}}{2n})\) As |x| < 1, when n goes to infinity, \(x^{n+2}\) approaches 0. So, the limit becomes: \(\lim_{n \to \infty} (\frac{x^2}{1} - 0)\)
05

Write the sum of the series

The sum of the convergent series when |x| < 1 is given by: \(\frac{x^2}{1}\)

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